zoj2112 主席树动态第k大 （主席树&&树状数组）
Dynamic Rankings
Time Limit: 10 Seconds Memory Limit: 32768 KB
The Company Dynamic Rankings has developed a new kind of computer that is no longer satisfied with the query like to simply find the k-th smallest number of the given N numbers. They have developed a more powerful system such that for N numbers a[1], a[2], ..., a[N], you can ask it like: what is the k-th smallest number of a[i], a[i+1], ..., a[j]? (For some i<=j, 0<k<=j+1-i that you have given to it). More powerful, you can even change the value of some a[i], and continue to query, all the same.
Your task is to write a program for this computer, which
- Reads N numbers from the input (1 <= N <= 50,000)
-
Processes M instructions of the input (1 <= M <= 10,000). These
instructions include querying the k-th smallest number of a[i], a[i+1],
..., a[j] and change some a[i] to t.
Input
The first line of the input is a
single number X (0 < X <= 4), the number of the test cases of the
input. Then X blocks each represent a single test case.
The first
line of each block contains two integers N and M, representing N
numbers and M instruction. It is followed by N lines. The (i+1)-th line
represents the number a[i]. Then M lines that is in the following format
Q i j k or
C i t
It
represents to query the k-th number of a[i], a[i+1], ..., a[j] and
change some a[i] to t, respectively. It is guaranteed that at any time
of the operation. Any number a[i] is a non-negative integer that is less
than 1,000,000,000.
There're NO breakline between two continuous test cases.
Output
For each querying operation,
output one integer to represent the result. (i.e. the k-th smallest
number of a[i], a[i+1],..., a[j])
There're NO breakline between two continuous test cases.
Sample Input
2
5 3
3 2 1 4 7
Q 1 4 3
C 2 6
Q 2 5 3
5 3
3 2 1 4 7
Q 1 4 3
C 2 6
Q 2 5 3
Sample Output
3
6
3
6
主席树太神了。
这题是动态第k大。
如果是不修改，直接主席树就可以了。
要修改要套如树状数组求和。
/* ***********************************************
Author :kuangbin
Created Time :2013-9-8 8:53:54
File Name :F:\2013ACM练习\专题学习\主席树\ZOJ2112.cpp
************************************************ */ #include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std; const int MAXN = ;
const int M = ;
int n,q,m,tot;
int a[MAXN], t[MAXN];
int T[MAXN], lson[M], rson[M],c[M];
int S[MAXN]; struct Query
{
int kind;
int l,r,k;
}query[]; void Init_hash(int k)
{
sort(t,t+k);
m = unique(t,t+k) - t;
}
int hash(int x)
{
return lower_bound(t,t+m,x)-t;
}
int build(int l,int r)
{
int root = tot++;
c[root] = ;
if(l != r)
{
int mid = (l+r)/;
lson[root] = build(l,mid);
rson[root] = build(mid+,r);
}
return root;
} int Insert(int root,int pos,int val)
{
int newroot = tot++, tmp = newroot;
int l = , r = m-;
c[newroot] = c[root] + val;
while(l < r)
{
int mid = (l+r)>>;
if(pos <= mid)
{
lson[newroot] = tot++; rson[newroot] = rson[root];
newroot = lson[newroot]; root = lson[root];
r = mid;
}
else
{
rson[newroot] = tot++; lson[newroot] = lson[root];
newroot = rson[newroot]; root = rson[root];
l = mid+;
}
c[newroot] = c[root] + val;
}
return tmp;
} int lowbit(int x)
{
return x&(-x);
}
int use[MAXN];
void add(int x,int pos,int val)
{
while(x <= n)
{
S[x] = Insert(S[x],pos,val);
x += lowbit(x);
}
}
int sum(int x)
{
int ret = ;
while(x > )
{
ret += c[lson[use[x]]];
x -= lowbit(x);
}
return ret;
}
int Query(int left,int right,int k)
{
int left_root = T[left-];
int right_root = T[right];
int l = , r = m-;
for(int i = left-;i;i -= lowbit(i)) use[i] = S[i];
for(int i = right;i ;i -= lowbit(i)) use[i] = S[i];
while(l < r)
{
int mid = (l+r)/;
int tmp = sum(right) - sum(left-) + c[lson[right_root]] - c[lson[left_root]];
if(tmp >= k)
{
r = mid;
for(int i = left-; i ;i -= lowbit(i))
use[i] = lson[use[i]];
for(int i = right; i; i -= lowbit(i))
use[i] = lson[use[i]];
left_root = lson[left_root];
right_root = lson[right_root];
}
else
{
l = mid+;
k -= tmp;
for(int i = left-; i;i -= lowbit(i))
use[i] = rson[use[i]];
for(int i = right;i ;i -= lowbit(i))
use[i] = rson[use[i]];
left_root = rson[left_root];
right_root = rson[right_root];
}
}
return l;
}
void Modify(int x,int p,int d)
{
while(x <= n)
{
S[x] = Insert(S[x],p,d);
x += lowbit(x);
}
} int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
int Tcase;
scanf("%d",&Tcase);
while(Tcase--)
{
scanf("%d%d",&n,&q);
tot = ;
m = ;
for(int i = ;i <= n;i++)
{
scanf("%d",&a[i]);
t[m++] = a[i];
}
char op[];
for(int i = ;i < q;i++)
{
scanf("%s",op);
if(op[] == 'Q')
{
query[i].kind = ;
scanf("%d%d%d",&query[i].l,&query[i].r,&query[i].k);
}
else
{
query[i].kind = ;
scanf("%d%d",&query[i].l,&query[i].r);
t[m++] = query[i].r;
}
}
Init_hash(m);
T[] = build(,m-);
for(int i = ;i <= n;i++)
T[i] = Insert(T[i-],hash(a[i]),);
for(int i = ;i <= n;i++)
S[i] = T[];
for(int i = ;i < q;i++)
{
if(query[i].kind == )
printf("%d\n",t[Query(query[i].l,query[i].r,query[i].k)]);
else
{
Modify(query[i].l,hash(a[query[i].l]),-);
Modify(query[i].l,hash(query[i].r),);
a[query[i].l] = query[i].r;
}
}
}
return ;
}
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